# Thread: Group of order pq must have subgroups of order p and q.

1. Originally Posted by Swlabr
Chauchy's Theorem is used to prove that finite groups with elements of order a power of $p$ are of order $p^n$. However, in this case the elements are of order $p$. This is a simpler case, and we can use the formula,

$|HK|=|H|.|K|/|H \cap K|$

That is to say, if we look at the problem in my proof each element has order $p$. As each element has order $p$ it will generate $p-2$ other elements of order $p$ and so if we take a minimum generating set, $P_1, P_2, \ldots, P_n$, the intersection of each pair is trivial, $G=P_1P_2 \ldots P_n$ and the formula can be applied. Thus,

$|G|=|P_1|.|P_2| \ldots |P_n| = p^n \neq pq$

and we have the contradiction needed to salvage my earlier proof.

I hope that makes sense...

It should be noted that that formula has an elementary proof, and so I believe that my proof will count as "elementary". I hope.
Are you telling me that whenever I have subgroups $H_1,H_2, \ldots,H_m$ of G such that $H_i \cap H_j = (e)$ for $i \neq j, |H_1H_2 \ldots H_m| = |H_1||H_2| \ldots |H_m|$? This is certainly true for $m=2$. Could you sugest a proof for $m$ any positive integer?

2. Originally Posted by ENRIQUESTEFANINI
Are you telling me that whenever I have subgroups $H_1,H_2, \ldots,H_m$ of G such that $H_i \cap H_j = (e)$ for $i \neq j, |H_1H_2 \ldots H_m| = |H_1||H_2| \ldots |H_m|$? This is certainly true for $m=2$. Could you sugest a proof for $m$ any positive integer?
No, I am not. For instance, take the Klein 4-group, $V_4=C_2 \times C_2$. We have 4 copies of $C_2$ in this group with trivial pairwise intersections. However, their product is $V_4$, a group of order 4 not of order 8. We need two more things - that the subgroups are of prime order, and that they form a minimal generating set. That is to say, $H_1H_2 \ldots H_m=G$ but $H_1H_2 \ldots H_{i-1}H_{i+1} \ldots H_m \neq G$.

The flaw in the counter example is that they do not form a minimal generating set. We need prime order as we do not know that $AB \cap C = <1>$ just because $A \cap C = B \cap C = <1>$ even if we do have a minimum generating set.

As to the proof - think...what would you normally do when something is true for $m=2$ (actually - easier than that - it's true for $m=1$)? You will need the formula I gave though, as well as the two conditions I've given, above.

3. I have another question regarding such groups that I started to ponder but could not really get anywhere on...are all such groups soluble? If so, then we are done as either they are abelian and so we have that they are isomorphic to $C_{pq} \equiv C_p \times C_q$, or else the derived subgroup is a proper normal subggroup which we can quotient out and then apply the correspondence theorem (I believe we can anyway) to get the other subgroup.

However, I can't seem to prove that all such groups are soluble, but I can't find any counter-examples (admittedly my examples were far from many in number...emm, $A_6$ was all I could really think of). Clearly there are no non-abelian nilpotent examples, as their centres are trivial.

4. Originally Posted by Swlabr
I have another question regarding such groups that I started to ponder but could not really get anywhere on...are all such groups soluble?
Yes, they are all solvable. You can easily see this from Burnside's theorem but this is a complete overkill.
This special case can also be proved with the Sylow theorems.

5. Originally Posted by Swlabr
As to the proof - think...what would you normally do when something is true for $m=2$ (actually - easier than that - it's true for $m=1$)? You will need the formula I gave though, as well as the two conditions I've given, above.
I would use induction, of course. I think I have a proof, and it would run as follows: suppose the proposition to be true for groups admitting a decomposition in m subgroups (hypothesis) and that $G=H_1 \ldots H_{m+1}$ is also a decomposition satisfying the three conditions. Then $G=HH_{m+1}$ with $H=H_1 \ldots H_m$. Obviously $H$ satisfies the three conditions. Now, lets suppose for a moment that two things happen:
(a) H is a subgroup of G.
(b) $H \cap H_{m+1} = (e)$.
Then, by the hypothesis, as H is a group in its own right, $|H| = |H_1| \ldots |H_m|$. And
by applying the formula, I get $|G| = |H||H_{m+1}|$ and the proposition is proved. Unfortunately, I don't see how (a) and (b) can be proved.

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